首先，我不确定这是提出这类问题的好地方。 如果没有，请告诉我在哪里问，我会删除它。
但是，由于我没有发现任何解决我问题的帖子，我的问题和下面的代码可能对其他人有用......
代码部分有点长，虽然我试图保持最小。但是，这部分应该只是为了表明我已经做了一些研究，而且我正在寻找更好的东西。
问题：
从x值和y值列表中，我想找到（或猜测）对应于给定y值的x值。
当由X和Y值定义的曲线是单调的时，我可以简单地插入函数x = f(y)：
import numpy as np
from scipy import interpolate
class AxisCam:
def __init__(self, x=None, y=None):
self.x = x if x else []
self.y = y if y else []
if len(self.x):
self.xMin = min(self.x)
self.xMax = max(self.x)
else:
self.xMin = None
self.xMax = None
if len(self.y):
self.yMin = min(self.y)
self.yMax = max(self.y)
else:
self.yMin = None
self.yMax = None
self._interpolX, self._interpolY = self.setInterpolator()
def setInterpolator(self, interpolator=interpolate.interp1d):
"""
Define the interpolator to use to approximate the axis cam positions
:param interpolator: interpolator function to use, default is scipy.interpolate.interp1d
:return: a tuple with the interpolator functions for x and y values
"""
if len(self.x) <= 0 or len(self.y) <= 0:
return None, None
with np.errstate(divide='ignore', invalid='ignore'): # silent the warnings caused by the interpolator
self._interpolX = interpolator(self.y, self.x) # x = f(y)
self._interpolY = interpolator(self.x, self.y) # y = f(x)
return self._interpolX, self._interpolY
def getX(self, yValue):
"""
Return x-value corresponding to a y-value using the interpolator
:param yValue: y-value we want to know the corresponding x-value
:return: x-value corresponding to the given y-value
"""
if yValue < self.yMin:
raise ValueError("value should be greater than the minimum y-value")
elif yValue > self.yMax:
raise ValueError("value should be lesser than the maximum y-value")
return float(self._interpolX(yValue))
def getY(self, value):
"""
Return a y-value corresponding to a x-value using the interpolator
:param value: x-value we want to know the corresponding y-value
:return: the y-value corresponding to the given x-value
"""
if value < self.xMin:
raise ValueError("value should be greater than the minimum x-value")
elif value > self.xMax:
raise ValueError("value should be lesser than the maximum x-value")
return float(self._interpolY(value))
x = [0, 0.351906, 0.703812, 1.055718] # The 1024 values for X and Y can be retrieved here : https://pastebin.com/5eHsRjZ3
y = [0.0, 0.000306, 0.002419, 0.008111]
ac = AxisCam(x, y)
print(ac.getX(100)) # returns 30.124163768271398
但是，当曲线是非单调的时，我不能。提出了一个例外
ValueError：x必须严格增加
因此，现在，我使用下面的getMonotonicParts方法将曲线分割成单调部分，我可以在每个单调部分上插入函数x = f(y)。
import numpy as np
from scipy import interpolate
class AxisCam:
def __init__(self, x=None, y=None):
self.x = x if x else []
self.y = y if y else []
if len(self.y):
self.yMin = min(self.y)
self.yMax = max(self.y)
else:
self.yMin = None
self.yMax = None
self._monotonicParts = self.getMonotonicParts()
def getMonotonicParts(self, interpolator=interpolate.interp1d):
parts = []
prevY = None # will store the previous value of y to compare with at each iteration
startIdx = None # will store the index of self.x and self.y where the monotonic part start from
direction = 0 # 0: Unknown - 1 : constant - 2: ascending - 3: descending
lenY = len(self.y)
for i, (x, y) in enumerate(zip(self.x, self.y)):
if prevY is None:
prevY = y
if startIdx is None:
startIdx = i
prevDir = direction
direction = 1 if y == prevY else 2 if y > prevY else 3
if prevDir != 0 and prevDir != direction: # Direction has changed => we have a new monotonic part
endIdx = i - 1
if direction == 3: # y values are increasing => we can interpolate on it
interp_func = interpolator(self.y[startIdx:endIdx], self.x[startIdx:endIdx])
elif direction == 1: # y values are decreasing => we need to reverse it to interpolate on it
xValues = self.x[startIdx:endIdx]
xValues.reverse()
yValues = self.y[startIdx:endIdx]
yValues.reverse()
interp_func = interpolator(yValues, xValues)
else: # y values are the same on the range => return one of these
def interp_func(value): return self.y[startIdx]
parts.append({'start': startIdx,
'end': endIdx,
'x0': self.x[startIdx],
'y0': self.y[startIdx],
'x1': self.x[endIdx],
'y1': self.y[endIdx],
'interp': interp_func})
startIdx = i
elif i == lenY - 1: # Add element on the last iteration
endIdx = i
if direction == 2:
interp = interpolator(self.y[startIdx:endIdx], self.x[startIdx:endIdx])
else:
interp = None
parts.append({'start': startIdx,
'end': endIdx,
'x0': self.x[startIdx],
'y0': self.y[startIdx],
'x1': self.x[endIdx],
'y1': self.y[endIdx],
'interp': interp})
prevY = y
return parts
def getX(self, yValue):
"""
Return a list of x-values corresponding to a y-value using the interpolator
:param yValue: y-value we want to know the corresponding x-value
:return: a list of x-values corresponding to the given y-value
"""
if yValue < self.yMin:
raise ValueError("value should be greater than the minimum y-value")
elif yValue > self.yMax:
raise ValueError("value should be lesser than the maximum y-value")
xValues = []
for part in self._monotonicParts:
if part['y0'] <= yValue <= part['y1'] or part['y0'] >= yValue >= part['y1']:
xValues.append(float(part['interp'](yValue)))
return xValues
x = [] # The 1024 values for X and Y can be retrieved here : https://pastebin.com/SL9RYYxY
y = [] # /!\ It is not the same values that the previous example /!\
ac = AxisCam(x, y)
print(ac.getX(100)) # returns [122.96996037206237, 207.6239552142487]
我的解决方案效果很好，但对我来说似乎有点牵强，我想知道是否还有其他更好的方法可以做到这一点。
我不知道做这种多插值的任何标准程序。但是如果你想要扩展它，你应该重构你的代码以使用numpy提供的所有内容。例如，你可以这样做：
import numpy as np
from scipy.interpolate import interp1d
# convert data lists to arrays
x, y = np.array(x), np.array(y)
# sort x and y by x value
order = np.argsort(x)
xsort, ysort = x[order], y[order]
# compute indices of points where y changes direction
ydirection = np.sign(np.diff(ysort))
changepoints = 1 + np.where(np.diff(ydirection) != 0)[0]
# find groups of x and y within which y is monotonic
xgroups = np.split(xsort, changepoints)
ygroups = np.split(ysort, changepoints)
interps = [interp1d(y, x, bounds_error=False) for y, x in zip(ygroups, xgroups)]
# interpolate all y values
yval = 100
xvals = np.array([interp(yval) for interp in interps])
print(xvals)
# array([ nan, 122.96996037, 207.62395521, nan])
这里nan表示超出范围的值（此算法将重复值视为单独的组）。